Question

A body is initially at rest, starts moving with a constant acceleration $2m{s^{ - 2}}$. Calculate the distance traveled in $5s$.
A) $10m$
B) $15m$
C) $25m$
D) $50m$

Answer 1

Hint: In kinematics, there are three equations of motion for a uniform acceleration. Now using the second equation of motion and substituting the given values. We have to find the distance traveled.
Velocity: it is defined as the rate of change of the object’s position with respect to the time. In other words, it can be defined as the displacement of the object in unit time. It is a vector quantity. The S.I unit is meter per second.
Acceleration: It is defined as the rate of change of velocity with respect to time. It is also a vector quantity. Two types of acceleration are uniform acceleration and non-uniform acceleration.
Uniform Acceleration: In simple words, it means that the acceleration is constant and neither increasing nor decreasing. Uniform acceleration is a change of equal velocity in equal intervals of time.

Formula Used:
Second Equation of Motion $s = ut + \dfrac{1}{2}a{t^2}$
Where $s$- Displacement
$u$- Initial velocity
$t$-Time
$a$- Acceleration

Complete step by step solution:
The displacement of a moving object is directly proportional to both velocity and time.
Initial velocity, $u = 0$
Acceleration, $a = 2m{s^{ - 2}}$
Time, $t = 5s$
Distance traveled, $s = ut + \dfrac{1}{2}a{t^2}$
Substituting the given values in this equation, we get
$s = (0 \times 5) + (\dfrac{1}{2} \times 2 \times {5^2}) = 25m$
The distance traveled in $5s$ is $25m$.

Answer is option $(C)$, $25m$.

Note: In kinematics, the equation of motion provides the concept of motion of an object such as velocity, position, time, speed, distance, and acceleration.
From the three equation of motion:
When acceleration is constant, the velocity is directly proportional to time and displacement is proportional to the square of time and combining these two things, displacement is proportional to the square of the velocity.